# Exact distribution of the MLE of the quantile function for the exponential distribution

Considering the pdf of the exponential distribution $$f(x,\theta) = \theta e^{-\theta x}$$ with $$x>0$$, and parameter $$\theta > 0$$. It is straightforward to find that the (biased) maximum likelihood estimator of $$\hat{\theta}$$ equals

$$\hat{\theta} = \frac{n}{\sum_{i = 1}^n x_i}.$$

Moreover, the quantile function of the distribution is equal to:

$$q_p = - \frac{\ln (1 - p)}{ \theta}$$

Here, at last, comes the question: is it possible for me to write the MLE of the quantile distribution $$\hat{q}_p$$ as: $$\hat{q}_p = - \frac{\ln(1-p)}{\hat{\theta}} ?$$ Moreover, how can I determine the exact distribution of the MLE of the quantile function? My intuition tells me that it should be normal, with mean equal to $$q_p$$ and variance equal to the inverse of the Fisher information, i.e.: $$\sqrt{n}\hat{q}_p \to^d \mathcal{N}(q_p, I^{-1})$$

Am I correct? Thanks in advance to everyone willing to argue.

• The first result is the functional invariance of MLE en.wikipedia.org/wiki/… Actually in your quantile MLE, the reciprocal of $\hat{\theta}$ results in a gamma distribution - the sum of i.i.d. exponential, and this is exact. The one you quote is the limiting distribution.
– BGM
Sep 13, 2019 at 7:41

is it possible for me to write the MLE of the quantile distribution $$\hat{q}_p$$ as
Yes, if you have some bijective transformation $$\tau = \tau(\theta)$$, then the MLE for $$\tau$$ is $$\hat \tau = \tau (\hat \theta)$$, where $$\hat \theta$$ is the MLE for $$\theta$$; this is called invariance property of MLE (the comment by @BGM points to a stronger version of it).
Hint: $$\hat{q}_p$$ is proportional to $$n/\hat \theta = \sum_{i=1}^n X_i\simeq \Gamma(n,\theta)$$.